OCR GCSE Chemistry A: Equilibria (J248) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

The questions below target C5.3 Equilibria in the J248 specification for OCR GCSE (9–1) Chemistry A (Gateway Science), Version 4.0, August 2026. Between them they test learning outcomes C5.3a–C5.3c and mathematical learning outcomes CM5.3i–CM5.3iv. Questions 1 to 7 (C5.3a, C5.3b and the maths) suit both tiers. Questions 8 to 12 apply Le Chatelier’s principle (C5.3c) and are labelled Higher Tier only.

Before you start, read the Equilibria study guide or the Equilibria revision notes. The OCR GCSE Chemistry hub and the J248 checklist show where this sub-topic fits.

How to use this set. Give a unit with every rate. For “Explain” questions on shifts, name the direction and say what happens to the substance in the question.

Questions

1. Anhydrous cobalt(II) chloride paper is used to test for water.

hydrated cobalt(II) chloride ⇌ anhydrous cobalt(II) chloride + water

(a) What does the symbol ⇌ show? [1] (b) A drop of a colourless liquid turns blue cobalt(II) chloride paper pink. What does this show about the liquid? [1] (c) Suggest how the pink paper could be turned blue again. [1]

2. State the type of system needed for a dynamic equilibrium, and two features of a reaction at dynamic equilibrium. [3]

3. A student writes: “At equilibrium, the reactants and products have equal concentrations.” Explain why the student is wrong. [2]

4. Gas C changes into gas D in a sealed container: C(g) ⇌ D(g). At the start the concentration of C is 0.50 mol/dm³ and there is no D.

Time (s)	0	5	10	15	20	25	30	35
[D] (mol/dm³)	0.00	0.21	0.31	0.36	0.39	0.40	0.40	0.40

(a) Calculate the mean rate of formation of D in the first 5 s. [2] (b) Calculate the mean rate between 5 s and 15 s and explain why it is lower than your answer to (a). [2] (c) Use the data to estimate when equilibrium is reached. [1] (d) Calculate the ratio of D to C at equilibrium, and the percentage of C converted to D. [2]

5. A graph of the data in question 4 is drawn. A tangent to the curve at 10 s passes through (0 s, 0.17 mol/dm³) and (25 s, 0.52 mol/dm³).

(a) Calculate the rate of formation of D at 10 s. [2] (b) State the gradient of a tangent drawn at 30 s and what it tells you. [1]

6. Describe how the rates of the forward and reverse reactions in question 4 change from the start until equilibrium is reached. [3]

7. In question 4, the forward rate is directly proportional to the concentration of C. At the start it is 0.060 mol/dm³/s.

(a) Calculate the forward rate at equilibrium. [2] (b) State the rate of the reverse reaction at equilibrium. Give a reason. [1]

8. (Higher Tier only) A U-tube contains this equilibrium: ICl(l) + Cl₂(g) ⇌ ICl₃(s). ICl is a brown liquid and ICl₃ is a yellow solid. The forward reaction is exothermic.

(a) More chlorine is passed into the tube. Predict what you would see and explain why. [2] (b) The U-tube is then put into a beaker of hot water. Predict what you would see and explain why. [2]

9. (Higher Tier only) For each reaction, say which way (if at all) the position moves when the pressure is raised.

(a) C(s) + CO₂(g) ⇌ 2CO(g). Explain your answer. [2] (b) N₂(g) + O₂(g) ⇌ 2NO(g) [1] (c) 2E(g) ⇌ F(g) + 3G(g) [1]

10. (Higher Tier only) In a sealed vessel, gas J forms gas K: 2J(g) ⇌ K(g). The percentage of K once equilibrium is reached was measured under four sets of conditions.

Temperature	Low pressure	High pressure
250 °C	30% K	55% K
450 °C	18% K	38% K

(a) Compare the two pressures. Explain how the results fit the equation. [2] (b) What do the results show about the forward reaction’s energy change? Explain. [2]

11. (Higher Tier only) Hydrogen can be made by reacting methane with steam:

CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g) forward reaction endothermic

(a) Predict the effect of raising the temperature on the amount of hydrogen at equilibrium. Explain your answer. [2] (b) Predict the effect of raising the pressure on the amount of hydrogen at equilibrium. Explain your answer. [2] (c) Suggest two other changes that would increase the amount of hydrogen produced. [2] (d) After a change, the system settles at a new equilibrium. Compare the forward and reverse rates now. [1]

12. (Higher Tier only) Iron(III) ions react with thiocyanate ions in solution:

Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) yellow colourless blood red

Plan an experiment to test the prediction that adding more of either reactant makes the mixture a deeper red. Include the apparatus, how you make it a fair test, what you record, and the result you expect, with a reason. [6]

Answers

1. (a) The reaction is reversible: it can go in both directions [1]. (b) The liquid contains water [1]. (c) Heat the paper (to drive off the water) [1]. Examiner insight: In (b), “the liquid is water” is weaker than “contains water”; a pink colour shows water is present, not that the liquid is pure.

2. A closed system [1]. Forward rate = reverse rate [1]. No concentration changes any more (each is constant) [1]. Examiner insight: “Rates are equal” needs both reactions named; “the reaction has the same rate” does not show you know two directions are involved.

3. At equilibrium the concentrations are constant, not necessarily equal [1]. They stay the same because the forward and reverse rates are equal, and the mixture can contain more product than reactant or the other way round [1]. Examiner insight: The second mark comes from linking constant concentrations to equal rates, not from repeating the word “constant”.

4. (a) mean rate = (0.21 − 0.00) ÷ 5 [1] = 0.042 mol/dm³/s [1]. (b) (0.36 − 0.21) ÷ 10 = 0.015 mol/dm³/s [1]. It is lower because there is less C left to react and the reverse reaction is faster as D builds up [1]. (c) At about 25 s, where [D] stops changing [1]. (d) [C] = 0.50 − 0.40 = 0.10 mol/dm³, so D : C = 4 : 1 [1]. Percentage converted = 0.40 ÷ 0.50 × 100 = 80% [1]. Examiner insight: A rate with no unit, or one missing the “/s”, is unlikely to earn the final mark.

5. (a) gradient = (0.52 − 0.17) ÷ (25 − 0) [1] = 0.014 mol/dm³/s [1]. (b) Zero: there is no overall change, because the system is at equilibrium [1]. Examiner insight: Read tangent points from far apart on the line; points close together make the gradient less accurate.

6. The forward rate is highest at the start and falls as C is used up [1]. The reverse rate starts at zero and rises as D forms [1]. The two rates become equal and then stay constant [1]. Examiner insight: Each mark needs a reason as well as a direction; “the forward rate goes down” alone describes without explaining.

7. (a) [C] falls from 0.50 to 0.10 mol/dm³, which is one fifth [1]. Forward rate = 0.060 × (0.10 ÷ 0.50) = 0.012 mol/dm³/s [1]. (b) 0.012 mol/dm³/s, because at equilibrium the forward and reverse rates are equal [1]. Examiner insight: Show the scale factor; a bare answer to a proportionality question gives the examiner nothing to credit if the arithmetic slips.

8. (a) More yellow solid forms (the brown liquid decreases) [1]. Adding a reactant shifts the position of equilibrium to the right to oppose the change [1]. (b) The yellow solid turns back into brown liquid [1]. A rise in temperature shifts the position in the endothermic direction, which is the reverse reaction [1]. Examiner insight: Observation questions need colours or states; “the equilibrium shifts left” is a reason, not an observation.

9. (a) Shifts to the left [1]. Only gases count: 1 gas molecule on the left and 2 on the right, so the position moves to the side with fewer gas molecules [1]. (b) No effect: 2 gas molecules on each side [1]. (c) Shifts to the left: 2 gas molecules on the left, 4 on the right [1]. Examiner insight: Counting carbon as a molecule in (a) gives 2 against 2 and the wrong answer; check state symbols first.

10. (a) At both temperatures there is more K at high pressure (30% to 55% at 250 °C; 18% to 38% at 450 °C) [1]. A higher pressure favours the side with fewer gas molecules, and K’s side has 1 against 2 for J [1]. (b) The forward reaction is exothermic [1]: at either pressure the percentage of K falls when the temperature goes up (for example 55% to 38%), so heating favours the reverse reaction, which must be endothermic [1]. Examiner insight: Quote a pair of values in which only one condition changes; comparing 30% at 250 °C with 38% at 450 °C mixes two variables.

11. (a) The amount of hydrogen increases [1], because the position shifts in the endothermic direction, which is the forward reaction [1]. (b) The amount of hydrogen decreases [1], because the left side has 2 gas molecules and the right side has 4, so higher pressure shifts the position to the left [1]. (c) Any two: use excess steam (or methane) [1]; remove hydrogen as it forms [1]. (d) They are equal again [1]. Examiner insight: In (b), the molecule count is the evidence; “pressure shifts it left” without the numbers 2 and 4 is unlikely to get the second mark.

12. Mix dilute iron(III) solution and dilute thiocyanate solution to make a red mixture [1]. Pour equal volumes into test tubes using a measuring cylinder [1]. Add a few drops of iron(III) solution to one tube and the same number of drops of thiocyanate solution to another [1]. Keep one tube as a control, adding the same volume of water so all tubes are diluted equally [1]. Compare the colours against a white tile and record them in a table; wear eye protection [1]. Expected result: both tubes with extra reactant turn deeper red, because the position shifts right to oppose the added reactant, making more FeSCN²⁺ [1]. Examiner insight: A plan gains credit for specific steps; “do a fair test” earns nothing unless you say which variables you keep the same.

Where marks are usually lost

• Writing “equal concentrations” instead of “constant concentrations” at equilibrium.
• Forgetting “closed system” when asked for the conditions for equilibrium.
• Giving a rate without units, or with the wrong ones.
• Using two table values when the question asks for the rate at one time, which needs a tangent.
• Counting solids, liquids or atoms in a pressure question.
• Stating a shift without saying what happens to the named substance.
• Linking a temperature rise to the exothermic direction.
• In plans, leaving out the control or the step that keeps dilution the same.

Next steps

• Recap the key words in the Equilibria revision notes.
• Go back to the Equilibria study guide for anything that went wrong.
• Mark C5.3 as done on the J248 checklist, then pick your next sub-topic from the OCR GCSE Chemistry hub.
• Find your next weak spot with a free 10-minute diagnostic.
• Book a free trial class.

Official syllabus

OCR GCSE (9–1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026), for first assessment in 2018, published by OCR. Sub-topic C5.3 Equilibria, learning outcomes C5.3a–C5.3c and mathematical learning outcomes CM5.3i–CM5.3iv.